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.Suppose that the health of a person could be at risk if theconcentration of vitamin D in the blood stream is less than 70.0nM/L (say 69.999 nM/L). Use Excel to cod...

Question

.Suppose that the health of a person could be at risk if theconcentration of vitamin D in the blood stream is less than 70.0nM/L (say 69.999 nM/L). Use Excel to code the datain column Sample 1 and put the result in column Sample_p. Codeconcentrations of 69.999 nM/L or lower as “1” and codeconcentrations of 70 and above as “0”. Calculatemanually a 99% confidence interval for the populationproportion for concentration of vitamin D less than 70.0 nM/L andconfirm your calculations using Exce

.Suppose that the health of a person could be at risk if the concentration of vitamin D in the blood stream is less than 70.0 nM/L (say 69.999 nM/L). Use Excel to code the data in column Sample 1 and put the result in column Sample_p. Code concentrations of 69.999 nM/L or lower as “1” and code concentrations of 70 and above as “0”. Calculate manually a 99% confidence interval for the population proportion for concentration of vitamin D less than 70.0 nM/L and confirm your calculations using Excel (use the normal approximation). Sample 1 Sample_p 87.076 69.03 74.436 84.161 88.414 94.052 97.15 65.791 96.522 71.736 67.113 75.427 82.938 93.804 87.749 84.458 85.309 60.938 83.53 112.029 76.55 80.752 105.515 73.447 87.211



Answers

Let $x$ be a random variable that represents red blood cell (RBC) count in millions of cells per cubic millimeter of whole blood. Then $x$ has a distribution that is approximately normal. For the population of healthy female adults, the mean of the $x$ distribution is about 4.8 (based on information from Diagnostic Tests with Nursing Implications, Springhouse Corporation). Suppose that a female patient has taken six laboratory blood tests over the past several months and that the RBC count data sent to the patient's doctor are $$4.9 \quad 4.2 \quad 4.5 \quad 4.1 \quad 4.4 \quad 4.3$$
i. Use a calculator with sample mean and sample standard deviation keys to verify that $\bar{x}=4.40$ and $s \approx 0.28$ ii. Do the given data indicate that the population mean RBC count for this patient is lower than $4.8 ?$ Use $\alpha=0.05.$

Right, where will the population has mean new equals 14 From the following sample data, we want to calculate the sample mean X bar and the sample standard deviation S. To do so let's remember the definition of these terms X bar Is the some of the data divided by n or in this case 15.1, I'm a sample standard deviation S is the sum of deviations about the mean square divided by n minus one or 2.51 Next we want to implement a right tail test. That is we want to test whether or not the population mean should actually be greater than the no mean 14. With us using the sample data with a significance level alpha equals 140.1 Where we are noted that X is approximately normally distributed. So to implement this test, we have to answer the following questions in order First, what is the significance of hypotheses? We've alpha equals 0.1 H not is new equals 14 H. A. Is that I mean is greater than 14. What distribution will use computer associated test statistic? Since the population standard deviation sigma is unknown. We have to use a student's T distribution which we know is okay to use because the shape of the distribution is normal, which is both symmetrical amount shaped from this. We calculate the T stat Which is given by this formula and reduces down to 1.386 for this problem? Next compute the p interval and sketch it out so we have degree of freedom and minus 29. We use the one tailed T. Table to identify that this tea interval falls between a p interval of 10.75 point one. That is because our T statistic falls between associated T values for these p values. We can graph this as the area under the student's t distribution. To the right of our T stat 1.386 as is highlighted in yellow on the right. What can we conclude from this? Well, we can conclude that P is greater than alpha, so we have statistically insignificant findings and we fail to reject astronaut, which means that we lack sufficient evidence that suggests our population mean is greater than the No. Mean 14.

We're going to be considering the 95% confidence interval for the difference between two population means so new one represents the I mean Huma Huma Glavine levels for women and Mewtwo represents the mean Hugh model being levels for men. So in the first part of the problem, we're going to be telling what the confidence interval suggests about the quality off. He mean him a global level in women on the mean hemoglobin level in men. So when we analyze this difference, we notice that the difference ranges from negative 1.76 grounds part the sea litter and negative 1.62 g per deciliter. So the difference is negative. And when the difference is negative, it means that mewtwo is greater than new one in other ones. This suggests that men have higher hemoglobin levels than women. That is on average. Next, uh, impact be we're going to be writing a brief statement that inter prints that confidence interval that we have here. So the confidence interval. This is a 95% confidence interval. So we can say that we are 95% confidence that live limits off negative 1.76 on negative 1.62 actually do contain the difference between them to population. Hemoglobin level means So that is what this means because it's a 95% confidence interval. So we're confident that the limits will be negative. 1.76 and negative one points. Uh, 62 Father will notice that this'll limits do not contain then zero did not contain zero. Okay. And because the limits do not contain zero, then the confidence interval suggests that there is a significant difference between the mean hemoglobin levels for women on the mean hemoglobin levels for men. So that difference is significant. So that brief statement he's as follows. We are 95% confident. Fact the limits off negative 1.76 on negative 1.62 actually do contain the difference. No, between the two population. Hey, more gripping level means that is for men and women. And we also say that because confidence interval does not contain zero, there is a significant difference. Okay, between mean hemoglobin levels for women and mean block him a clipping levels for men. So next, in part C of the exercise, we're going to express this confidence interval when the mission is off, men are given by population one on the measures from women are given by population to. In other words, if we stop this around so that me one represents men and Mewtwo represents women. So if that was the case, then the differences will do no longer be negative. The differences would now be positive and therefore we would have the following confidence interval. New ones minus new to would have the same difference, but this time positive so that the difference would be positive. In other words, we would have positive 1.62 to be the smaller difference and positive 1.76 to be their greater difference. So that would be the confidence interval, with measures from men being population one and measures from women being population too.

So we're given a distribution for red blood cell count, and For saying that the average red blood cell count is 4.8 and that a standard deviation is .3. So let's just go out a little bit and here's 5.1, Here's 5.4, here's 5.7. And if we subtract away .3 we're at 4.5, we're at 4.2, we're at 3.9 And we want to convert these into Z values. So what's the likelihood? 4.5 is less than X. And So if we take our we know our Z score is equal to our score, minus the mean divided by the standard deviation. So converting this, we would take that 4.5 minus the mean, Divided by a standard deviation. And we can actually see that right there on the uh score right here, that 4.5 was corresponding to a negative one, part b Um part be asked us, what about 4.2? If I have X Greater than excuse me? Less than 4.2. Well we can see that that corresponds to a value of the Z. score being less than negative two Part C. If we want to find the probability of being between four And 5.5. Yeah. And we use this conversion so basically I'm going to substitute by foreign here And so we have the four -4.8 Divided by the .3 And that's going to come out to approximately negative 2.67. Mhm. Now that's a Z value and then likewise substitute this in here. And So change that to a 5.5 And I'll have to do a little insert 5.5 And that gives me a value of positive 2.33. And again if I look at where 5.5 would be about right here, that makes sense. That that's going to be a score of 2.33, and likewise, of four, is going to be right here. And that value of negative 2.67 looks correct. So let's keep going now, we want to convert these into a score, we know the Z score is a negative 1.44, And if the Z score is negative 1.44, that means our Z score is going to be negative 1.44, It's going to be the score minus the mean, which is the 4.8 divided by the standard deviation, which will come out to be, I'll take my 4.8 and then it will be plus, and then that is a negative Z score times .3, and that will be what my calculation is. So I'm just going to type that in one time and then we can just do a whole bunch of uh Plus .3 times negative 1.44. And that score would correspond with Having a value up, whoops. It's not it's an X core. now of 4.368 E. We want 1.28 to be higher than Z. Well, again, that score will end up being 4.8 plus 1.28 Standard deviations higher. So I can carol back up And change my Z score or my standardized score to positive 1.28. And that corresponds with the value of uh 5.184 is less than that value. Yeah, We want between negative 2.25 for a Z score and negative one. Well, we know what negative one is, -1 is at 4.5. So that will save us somewhere. Yeah and Now we just have to convert this one and that's going to be that 4.8 Plus negative 2.25 times 0.3 and I can't go back up again and just change that value To a -2.25. And that comes out to a 4.125. And then lastly hope I'm not right yep. The last one is G. And it said if you had a Red blood cell count that was greater than or equal to 5.9, would that be high? Oh and yes Three standard deviations is 5.7. So we would be way up here. So this would be very unusual. Yes, that would be unusual. Or there could be some type of a problem.

11 Good chili on that it is called is equal toe. The X value one is the me over understanding deviation which, approximately equal to negative one So negative one is one of than is one of than that for which and for which and be, um does the score is equal to the X value minus the mean over the standard Division, which actually is equal to a negative to so that that many here is is smaller than negative for seen. Uh, depending here is four minus four point 8/1 4.0.3, which is negative, 2.67 and then for the other many is 5.54 point eight over opened three. She's a toe. So this is where you would like between negative 2.67 and 2.33 for preaching the, um, the X very equal to the mean minus 1.44 time open three, which is 4.36 eight. So the expert is smaller than 4.36 eight. For questions in me, the X daily is equal toe me 4.8 plus, uh, doesn't the score multiplied by Son of deviation, which is equal to 5.18 four. So 5.18 for is smaller than X. The X value is equal toe the mean minus 2.2 15 times 4.3, which is 4.125 and it's ready for other friend is 4.8 using one. The meaning is one times open three, which is 4.5. So 4.1 toe Find explosives between these two values. Question regime. So this the score is equal toe X radio, minus the mean over the summer, the energy 3.6 seven. So the usual values are written to something division off, I mean, and thus have this call between negative 70.2. So that is 3.6. So it's not just than two, so 5.1 is unusual number.


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